
The period of a function \[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\] (where \[\left\{ x \right\}\]denotes the fractional part of x) is
A.2
B.1
C.3
D.None of these
Answer
577.8k+ views
Hint: In this question first of all we calculate the period of all the individual terms of the given function. After that, to calculate the period of a function take Least common multiple (LCM) of period of all the terms. Which will be the period of our given function.
Complete step-by-step answer:
We have given \[f\left( x \right) = \]\[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\]
As we know that,
Period of fractional part is always 1
\[ \Rightarrow x = 1\]
\[ \Rightarrow \]Period of \[{2^{\left\{ x \right\}}}\]= 1 \[ \ldots \left( 1 \right)\]
For period of \[{3^{\left\{ {\dfrac{x}{2}} \right\}}}\]
Consider, \[\dfrac{x}{2} = 1\]
\[ \Rightarrow \]\[x = 2\]
\[ \Rightarrow \]Period of \[{3^{\left\{ {\dfrac{x}{2}} \right\}}}\]=2 \[ \ldots \left( 2 \right)\]
Period of sine function is \[2\pi \]
\[ \Rightarrow \] Period of \[\sin \pi x\]=\[2\pi \]
\[ \Rightarrow \] Period of \[\pi x = 2\pi \]
\[ \Rightarrow x = \dfrac{{2\pi }}{\pi }\]
\[ \Rightarrow \] Period of \[\sin \pi x\] is \[x = 2\] \[ \ldots \left( 3 \right)\]
Period of cosine function is \[2\pi \]
For period of \[\cos 2\pi x\]
Consider, \[2\pi x = 2\pi \]
\[ \Rightarrow x = \dfrac{{2\pi }}{{2\pi }}\]
\[ \Rightarrow x = 1\] \[ \ldots \left( 4 \right)\]
For calculating the period of any function we have to take LCM of the period of all the terms given in the above function.
So, period of \[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\] is LCM \[\left( {1,2,2,1} \right)\] \[ \ldots \](from 1,2,3 and 4)
Thus, period of \[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\] is 2.
Hence, option A. 2 is the correct option.
Note: Period of a function: The distance between the repetition of any function is called the period of the function.
For a trigonometric function the length of one complete cycle is called a period.
Fraction part function is a periodic function with period 1.
Complete step-by-step answer:
We have given \[f\left( x \right) = \]\[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\]
As we know that,
Period of fractional part is always 1
\[ \Rightarrow x = 1\]
\[ \Rightarrow \]Period of \[{2^{\left\{ x \right\}}}\]= 1 \[ \ldots \left( 1 \right)\]
For period of \[{3^{\left\{ {\dfrac{x}{2}} \right\}}}\]
Consider, \[\dfrac{x}{2} = 1\]
\[ \Rightarrow \]\[x = 2\]
\[ \Rightarrow \]Period of \[{3^{\left\{ {\dfrac{x}{2}} \right\}}}\]=2 \[ \ldots \left( 2 \right)\]
Period of sine function is \[2\pi \]
\[ \Rightarrow \] Period of \[\sin \pi x\]=\[2\pi \]
\[ \Rightarrow \] Period of \[\pi x = 2\pi \]
\[ \Rightarrow x = \dfrac{{2\pi }}{\pi }\]
\[ \Rightarrow \] Period of \[\sin \pi x\] is \[x = 2\] \[ \ldots \left( 3 \right)\]
Period of cosine function is \[2\pi \]
For period of \[\cos 2\pi x\]
Consider, \[2\pi x = 2\pi \]
\[ \Rightarrow x = \dfrac{{2\pi }}{{2\pi }}\]
\[ \Rightarrow x = 1\] \[ \ldots \left( 4 \right)\]
For calculating the period of any function we have to take LCM of the period of all the terms given in the above function.
So, period of \[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\] is LCM \[\left( {1,2,2,1} \right)\] \[ \ldots \](from 1,2,3 and 4)
Thus, period of \[{2^{\left\{ x \right\}}} + \sin \pi x + {3^{\left\{ {\dfrac{x}{2}} \right\}}} + \cos 2\pi x\] is 2.
Hence, option A. 2 is the correct option.
Note: Period of a function: The distance between the repetition of any function is called the period of the function.
For a trigonometric function the length of one complete cycle is called a period.
Fraction part function is a periodic function with period 1.
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